Question

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently, the mean collision time between the gas molecule changes from ${\tau }_1$ to ${\tau }_2$. If $\frac{C_p}{C_v}=\gamma$ for this gas then a good estimate for $\frac{{\tau }_2}{{\tau }_1}$ is given by

• A. $2$
• B. $\frac{1}{2}$
• C. ${\left(\frac{1}{2}\right)}^{\gamma }$
• D. ${\left(2\right)}^{\frac{\gamma +1}{2}}$

Answer 1

The correct option is D ${\left(2\right)}^{\frac{\gamma +1}{2}}$

From the expression of mean free path,

$l=\frac{1}{\sqrt{2}n\pi d^2}$

Where, $n\to$ number of molecules per unit volume,
$n=\frac{N}{V}$

$\Rightarrow \tau =\frac{l}{v}$

Where, $v=\sqrt{\frac{3RT}{M}}$

$\therefore \tau =\frac{V\sqrt{M}}{N\pi d^2\sqrt{6RT}}$

$\Rightarrow \tau \propto \frac{V}{\sqrt{T}}$ $...\left(\text{i}\right)$

Equation of adiabatic process is

$\Rightarrow T\propto \frac{1}{V^{\gamma -1}}$

$\Rightarrow \tau \propto V^{1+\frac{\gamma -1}{2}}\left(\text{using (i)}\right)$

$\Rightarrow \tau \propto V^{\frac{1+\gamma }{2}}$

$\Rightarrow \frac{{\tau }_2}{{\tau }_1}={\left(\frac{2V}{V}\right)}^{\frac{1+\gamma }{2}}=(2{)}^{\frac{1+\gamma }{2}}$

Hence, option $\left(D\right)$ is correct.